# How high can an object be ejected out of water?

BenL
You know when you push a basket ball in the water and let it go. It goes back to the surface very fast and is expulsed from the water and goes above the water surface.

Now take it to an extreme, you place a capsule of air at the bottom of the ocean and set it free, how high can it reach above the water surface? Could it be high enough to go into orbit? Or is there a cap to the speed of the ball into the water?

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You know when you push a basket ball in the water and let it go. It goes back to the surface very fast and is expulsed from the water and goes above the water surface.

Now take it to an extreme, you place a capsule of air at the bottom of the ocean and set it free, how high can it reach above the water surface? Could it be high enough to go into orbit? Or is there a cap to the speed of the ball into the water?
It would have to depend on the size and shape of the "capsule". A suitably shaped container, released at sufficient depth, would reach terminal velocity. Drag and buoyant force would govern that terminal velocity. In the absence of an actual calculation, I looked up the speed of underwater torpedos. That suggests possible speeds of around 10+m/s, which would produce a height of around 5m (equating gravitational PE with KE). Not very stunning. The height would be less than that estimate because I'm sure the buoyant force would be less than the force from torpedo propulsion.

russ_watters
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First of all, you need to be much more careful with your assumptions. What does it mean for you to "place a capsule of air at the bottom of the ocean". In the case of a ball, where do you fill it up? At the surface or at the bottom? If you fill it with surface pressure, it will get completely squished once you bring it down to the bottom of the ocean. If you fill it with the local pressure at the bottom of the ocean, it will explode long before it reaches the surface. It is not even completely clear that the ball will rise if it is filled with atmospheric pressure at the surface, since it will be completely squished before reaching the bottom. The density of the material that the ball is made of will play the major role in this particular issue.

However, for the sake of the argument, let us assume that you somehow manage to create a container that can withstand any kind of external pressure and you fill it with atmospheric pressure air. Once you release it from the bottom, it will start accelerating upwards, but the drag force from the water will quite quickly balance out the lift from the pressure gradient and it will reach terminal velocity. This is essentially the velocity it will exit the water with. Compare to dropping a penny from a skyscraper - it will not hit the ground with a speed that is ##v = \sqrt{2gh}## - it will reach a terminal velocity for which the drag force from the air balances out the gravitational acceleration.

Mentor
You know when you push a basket ball in the water and let it go. It goes back to the surface very fast and is expulsed from the water and goes above the water surface.

Now take it to an extreme, you place a capsule of air at the bottom of the ocean and set it free, how high can it reach above the water surface? Could it be high enough to go into orbit? Or is there a cap to the speed of the ball into the water?
It's no different than throwing the ball into the air with your hands - the height it reaches depends on how fast it's moving when it leaves your hands, or in this case the surface of the water.

That speed doesn't keep increasing as you start the ball deeper in the water. There's a balance between the upwards bouyancy force and the resistance from trying to push the ball faster through the water; the ball won't go any faster than the speed at which the these two forces balance.

I'd expect that you've already seen close to the maximum possible speed and height.

sophiecentaur
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Hello Ben,
To escape Earth (getting into orbit is a little difficult when you emerge vertically from the water) an escape velocity of 11186 m/s is needed. Over mach 7.5 . (7.5 x speed of sound)

The water definitely can't reach mach 1.

A more detailed calculation would find a terminal velocity closer to mach 0.001.

sophiecentaur
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Drag and buoyant force would govern that terminal velocity.

@sophiecentaur has it right. I would say the same thing as, "size and shape are all important in determining the answer."

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To escape Earth (getting into orbit is a little difficult when you emerge vertically from the water) an escape velocity of 11186 m/s is needed. Over mach 7.5 . (7.5 x speed of sound)
And that is neglecting air resistance ...

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once again, and keep the ideas coming !

lekh2003
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@sophiecentaur has it right. I would say the same thing as, "size and shape are all important in determining the answer."
I was thinking in terms of the effect of the vast bubbles of methane gas that get released from the seabed occasionally. The highest point that any of that methane would reach could be considerable because you could perhaps expect a smoke-ring type of turbulence. But that wouldn't count for your question, I guess.

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once again, and keep the ideas coming !
That's what happens when you ask an interesting question, Ben.

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I would like to point out that I lowered my body mass during the past year!

Either way, I think we all pretty much agree.

sophiecentaur
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I would like to point out that I lowered my body mass during the past year!
vast bubbles of methane gas

I hope the two are not correlated!

Orodruin
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If you fill it with the local pressure at the bottom of the ocean, it will explode long before it reaches the surface.
Maybe some kind of mechanism that let's out air? That would help to make the question workable.

This question is quite interesting, I'm not well versed in drag and fluid movement, what would the terminal velocity of something like a regular sized basketball look like?

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@lekh2003 Look at this link about Stoke's Law It will give you an idea of the speed of a sphere through a fluid for a given force. A sphere is a particularly common example of the use of Stoke's law. The buoyant force will just be the weight of the ball full of water. Making and solving that simple equation will tell you the terminal velocity. The values of all the unknowns are available.
The problem of getting the ball to 'escape' the clutches of the water surface is harder and would reduce the effective exit velocity. Have a go!

lekh2003
willem2
This question is quite interesting, I'm not well versed in drag and fluid movement, what would the terminal velocity of something like a regular sized basketball look like?
If I use the equation for turbulent drag and ignore the mass of the ball i get:
$$\frac{1}{2} \rho v^2 C_d \pi r^2 = \frac {4}{3} g \pi \rho r^3$$
ρ is the density of the water
Cd = 0.47 for a spherical shape.
r is the radius of the ball. (12 cm for a full-sized basketball, volume is 0.0072 m3 weight 0.62 kg)
The result is.
$$v^2 = \frac {8 r g} {3 C_d}$$
v is proportional to the square root of the ball radius.
I get v = 2.6 m/s from this. This would result in a height of only 0.33 m, and I've seen balls that were smaller than a basketball go higher.
It seems it is very important what happens at the surface. If the ball is half out of the water, it would still have 30 N of buoyancy, there would be no more drag, and it would accelerate at 30/0.62 = 49 ms-2. The drag formula is likely not valid close to the surface. A very light ball such as a ping-pong ball might accelerate much faster near the surface and go higher than a basket ball, even if its speed through the water at larger depths is slower.

What happens to the water in the wake of the ball is also critical.
According to this link, the ball often goes higher if released at a moderate depth (a few ball diameters ) than at larger depths.
At larger depths sideways movement is observed under water, probably due to vortex shedding, and the ball sometimes doesn't clear the water at all.
Water exit dynamics of buoyant spheres
Tadd T. Truscott, Brenden P. Epps, and Randy H. Munns
Phys. Rev. Fluids 1, 074501 – Published 1 November 2016
https://journals.aps.org/prfluids/abstract/10.1103/PhysRevFluids.1.074501

Behind a paywall, but you can still see graphs of the results and pictures of the ball popping out with the turbulence behind it.

lekh2003, BvU and Orodruin
According to this link, the ball often goes higher if released at a moderate depth (a few ball diameters ) than at larger depths.
I observed this too in a pool with a ~25cm diameter rubber ball. Maximum jump height was from less than 1 m depth.

It would be interesting to try this with a streamlined buoy.

lekh2003
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It would be interesting to try this with a streamlined buoy.
Or a penguin!

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